Q 5. - Intermediate Algebra

Question

In the following exercises, Solve the system of equation by graphing 

$$\begin{gathered} 3x+y=-3 \\ 2x+3y=5 \end{gathered}$$

Step-by-Step Solution

Answer

The solution of the system of linear equations is (-2,3).

Step 1. Given information

The given linear equations are,                        

$$\begin{gathered} 3x+y=-3 \\ 2x+3y=5 \end{gathered}$$

Step 2. Solving equations for the intersection points

First, solve both of these equations for y such that their slopes and y intercepts may be easily graphed.

And find the slope and y-intercept by solving the first equation for y

$$\begin{gathered} 3x+y=-3 \\ y=-3x-3 \\ \mathrm{Here} \mathrm{the} \mathrm{slope} \mathrm{is} \mathrm{m} = -3 \\ \mathrm{And} \mathrm{the} \mathrm{y}-\mathrm{intercept} \mathrm{is} b = -3 \end{gathered}$$

Again find the slope and y-intercept by solving the Second equation for y.  

$$\begin{gathered} 2x + 3y = 5 \\ 3y = - 2x + 5 \\ y = -\frac{2}{3} \times x + \frac{5}{3} \\ \mathrm{Here} \mathrm{the} \mathrm{slope} \text{i}s m = - \frac{2}{3} \\ \mathrm{And} \mathrm{the} \mathrm{y}-\mathrm{intercept} is b =\frac{5}{3} \end{gathered}$$

Step 3. Graph obtained

Step 4. Checking the values

$$\begin{gathered} \mathrm{Check}: \\ \mathrm{First} \mathrm{substitute} \mathrm{x}=-2, \\ \mathrm{y}=3 \\ \mathrm{into} \mathrm{the} \mathrm{equation} 3\mathrm{x}+\mathrm{y}=-3 \\ 3x+y=-3 \\ 3(-2)+3=-3 \\ \mathrm{Also} \mathrm{substitute} \mathrm{x}=-2, \mathrm{y}=3 \mathrm{into} \mathrm{the} \mathrm{equation} \\ 2x+3y=5 \\ 2x+3y=5 \\ 2(-2)+3\left(3\right)=5 \\ -4+9=5 \\ 5=5 \\ \mathrm{This} \mathrm{is} \mathrm{true}.\mathrm{Therefore},\mathrm{the} \mathrm{solution} \mathrm{of} \mathrm{the} \mathrm{system} \mathrm{of} \mathrm{linear} \mathrm{equations} \mathrm{is} \mathrm{that} \mathrm{boxed}(-2,3) \end{gathered}$$